Let $f(x, y) = 2x + 3y - 1$ and $g(t) = (t^2, 2t)$. $h(t) = f(g(t))$ $h'(4) = $
Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(4) = \nabla f(g(4)) \cdot g'(4)$. $\begin{aligned} &g(4) = (16, 8) \\ \\ &g'(4) = (2t, 2) = (8, 2) \\ \\ &\nabla f = (2, 3) \\ \\ &\nabla f(g(4)) = \nabla f(16, 8) = (2, 3) \end{aligned}$ Substituting: $h'(4) = (2, 3) \cdot (8, 2) = 16 + 6 = 22$ Answer $h'(4) = 22$